Problem: A particle moves in the $xy$ -plane so that at any time $t\geq 0$ its coordinates are $x=t^3-2t$ and $y=3t+1$. What is the particle's velocity vector at $t=3$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $(25,3)$ (Choice B) B $(14,4)$ (Choice C) C $(20,-10)$ (Choice D) D $(21,10)$
Solution: Background When solving motion problems, it's important to remember the relationship between the position vector $(x,y)$, the velocity vector $\vec{v}(t)$, and the acceleration vector $\vec{a}(t)$ of the moving particle: $\vec{v}(t)=\left(\dfrac{dx}{dt},\dfrac{dy}{dt}\right)$ $\vec{a}(t)=\dfrac{d}{dt}\vec{v}(t)=\left(\dfrac{d^2x}{dt^2},\dfrac{d^2y}{dt^2}\right)$ Setting up the math We are given that the particle's coordinates are $x=t^3-2t$ and $y=3t+1$, which means its position vector is $(t^3-2t, 3t+1)$. We are asked to find the particle's velocity vector at $t=3$. In other words, we need to find $\vec{v}(3)$. Finding $\vec{v}(t)$ $\begin{aligned} \vec{v}(t)&=\left(\dfrac{d}{dt}(t^3-2t),\dfrac{d}{dt}(3t+1)\right) \\\\ &=(3t^2-2,3) \end{aligned}$ Finding $\vec{v}(3)$ $\begin{aligned} \vec{v}({3})&=(3({3})^2-2,3) \\\\ &=(27-2,3) \\\\ &=(25,3) \end{aligned}$ In conclusion, the particle's velocity vector at $t=3$ is $(25,3)$.